Finding the change in Rotational Kinetic Energy

In summary, a sphere with a 3 kg * m^2 moment of inertia about an axis through its center has its angular velocity changed by 4 rad/s owing to a 2 N*m torque applied about that axis for 6s. The change in its rotational kinetic energy in J is either 144 J or -96 J, depending on whether the angular speed is increased or decreased. This information can be found by using the work-kinetic energy theorem and the definition of rotational work.
  • #1
crazyog
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0

Homework Statement


A sphere with a 3 kg * m^2 moment of inertia about an axis through its center has its angular velocity changed by 4 rad/s owing to a 2 N*m torque applied about that axis for 6s. If its initial angular velocity is 10 rad/s, the change in its rotational kinetic energy in J is...
The answer according to the book is 144 J.

Homework Equations


1/2Iiwi^2 - 1/2Ifwf^2 = change in Ke
T=I(a)



The Attempt at a Solution


The wording of the question is kind of throwing me off because i don't understand the part about the Torque.

Thanks
 
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  • #2
It will be helpful to you to know that there are analogies in rotational mechanics to the definitions and equations you learned in linear mechanics.

Back in linear mechanics, you have the definition of work:

W = F · delta_x

and the work-kinetic energy theorem:

W = KE_f - KE_i .

In rotational mechanics, the rotational work will now be

W_rot = (torque) · (delta_theta) ,

with delta_theta being the angle the object has turned through while the torque is applied.

The work-kinetic energy theorem still applies,
now with

KE_rot = (1/2) · I · (w^2) .
 
  • #3
Ok so the initial KE_rot = 1/2 * I * (w^2)
= 1/2 * 3 * (10^2) = 150
I see what you are saying that W_rot = torque * delta theta
but how would that apply I am not given an angle?
and am I finding the W_rot and subtracting it from 150?
 
  • #4
crazyog said:
A sphere with a 3 kg * m^2 moment of inertia about an axis through its center has its angular velocity changed by 4 rad/s … If its initial angular velocity is 10 rad/s, the change in its rotational kinetic energy in J is...
The answer according to the book is 144 J.

crazyog said:
… I am not given an angle?
and am I finding the W_rot and subtracting it from 150?

Hi crazyog! :smile:

You don't need an angle or a torque or anything else for this part of the question …

You know it goes from 10 rad/s to 14 rad/s … so what is the increase in rotational KE? :smile:
 
  • #5
Oh ok so when they say "changed by 4 rad/s" it means it is now 14 m/s
so that would be

1/2(3)(14^2) = 294 J
294 - 150 = 144 J

Thank you!
 
  • #6
tiny-tim said:
You don't need an angle or a torque or anything else for this part of the question …

Oh, yes, quite so! Since they gave that information, I thought we'd need it somewhere. (Presumably, there are other parts to this problem...)

You know it goes from 10 rad/s to 14 rad/s … so what is the increase in rotational KE? :smile:

I wonder how they expected the student to know that the angular speed was being increased. The problem states that the "angular velocity (is) changed by 4 rad/s". You get a different magnitude for the rotational KE change if the angular speed goes to 6 rad/sec...
 
  • #7
dynamicsolo said:
I wonder how they expected the student to know that the angular speed was being increased. The problem states that the "angular velocity (is) changed by 4 rad/s". You get a different magnitude for the rotational KE change if the angular speed goes to 6 rad/sec...

yes … you're right! … :smile:

Now I'm wondering why I assumed it was increasing! :redface:
 
  • #8
And the change in angular momentum (rotational impulse) is

delta_L = (torque)·(delta_t) .

So the change in angular momentum is

delta_L = +/- (2 N·m)·(6 sec) = +/-12 kg·(m^2)/sec ,

so the new angular momentum would be either

L' = (3 kg·m^2)·(10 rad/sec) + 12 kg·(m^2)/sec
= 42 kg·(m^2)/sec

or

L' = (3 kg·m^2)·(10 rad/sec) - 12 kg·(m^2)/sec
= 18 kg·(m^2)/sec ,

which are consistent with final angular speeds of 14 or 6 rad/sec .

So the other information in the problem doesn't provide a clue either...

So a second valid answer would be -96 J for the change in rotational KE. (I think the solver for the textbook must have assumed the speed was increased.)
 
Last edited:

What is rotational kinetic energy?

Rotational kinetic energy is the energy an object possesses due to its rotation around an axis. It is dependent on the object's moment of inertia and angular velocity.

How do you calculate the change in rotational kinetic energy?

The change in rotational kinetic energy can be calculated using the equation ΔK = ½I(ωf2 - ωi2), where ΔK is the change in kinetic energy, I is the moment of inertia, ωf is the final angular velocity, and ωi is the initial angular velocity.

What factors affect the change in rotational kinetic energy?

The change in rotational kinetic energy is affected by the object's moment of inertia, angular velocity, and the applied torque. Additionally, external forces such as friction or air resistance can also affect the change in rotational kinetic energy.

What are some real-world examples of changes in rotational kinetic energy?

A few examples of changes in rotational kinetic energy are a spinning top slowing down, a swinging pendulum, a rolling ball gaining speed as it moves down a hill, and a figure skater spinning faster by pulling in their arms.

How is rotational kinetic energy related to linear kinetic energy?

Rotational kinetic energy and linear kinetic energy are both forms of kinetic energy, but they are dependent on different factors. Rotational kinetic energy is dependent on an object's rotational motion, while linear kinetic energy is dependent on an object's linear motion. However, these energies can be converted into one another, such as when a rolling wheel turns its rotational energy into linear energy.

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